Alternatives to proving the uncountability of number between 0 and 1

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SUMMARY

The discussion centers on proving the uncountability of real numbers between 0 and 1, specifically addressing the limitations of finite decimal representations. A participant argues that Cantor's diagonal argument is necessary, as finite decimal representations lead to a countable set. The conversation highlights the bijection between the set of real numbers with finite decimal expansions and finite sets, ultimately concluding that the set of all real numbers in [0,1] with finite decimal representations is countable. The need for deeper understanding of mathematical concepts is emphasized throughout the exchange.

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  • Understanding of Cantor's diagonal argument
  • Familiarity with bijections in set theory
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  • Basic concepts of decimal representations in real numbers
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robertjford80
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I'm aware of Cantor's diagonal argument but can't you prove the uncountability of reals between 0 and 1 using a simpler method? For instance, take the number .1 sooner or later in order to get a list of the reals between 0 and 1, you're going to have to get to .2 but before you can get to .2 you have to write .11 on your list then before you can get to .12 you have to write .111. In other words if you're obligated to open up a new decimal place then you'll never get to .2
 
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Since there are only countably many reals with finite decimal representations, and since this observation only accounts for these representations, it will not suffice. Something more sophisticated like the diagonal argument is needed.
 
jgens said:
Since there are only countably many reals with finite decimal representations

I need more details as to what you mean.
 
robertjford80 said:
I need more details as to what you mean.

In the future try thinking about the claim for more than three minutes before asking to have it spelled out for you. In any case, notice that the set An of all real numbers in [0,1] whose decimal expansion has length n is in bijection with the set {0,...,9} x ... x {0,...,9} (this is an n-fold product). Since this latter set is finite, it follows that An is finite. Thus the set A of all real numbers in [0,1] with finite decimal representations is given by ∪An and since a countable union of countable sets is again countable the claim follows.
 
jgens said:
In the future try thinking about the claim for more than three minutes before asking to have it spelled out for you.
What a nice person you are. Got a love a man that insults people so easily.

In any case, notice that the set An of all real numbers in [0,1] whose decimal expansion has length n is in bijection with the set {0,...,9} x ... x {0,...,9} (this is an n-fold product).
You're going to have to restate this in informal English.

Since this latter set is finite
I'm not convinced that it's finite. Explain.
 
robertjford80 said:
What a nice person you are. Got a love a man that insults people so easily.

It was not an insult, it was a request. PF is not a place for other people to do the thinking for you. There is an expectation that you think something through before asking.

You're going to have to restate this in informal English.

It would do you good to familiarize yourself with mathspeak, but that point aside I am not sure which part of that sentence is tripping you up, so I need some indication on which parts need clarifying.

I'm not convinced that it's finite. Explain.

Ten minutes is also not a sufficient period of time. Think about the claim some more and if after a couple hours you cannot understand why the set {0,...,9} x ... x {0,...,9} (again an n-fold product here) is finite, then come back and ask.
 
The OP won't be coming back any time soon, so there's no point in keeping this open.
 

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